# NCERT solution class 12 chapter 1 Integrals exercise 1.5 mathematics part 2

## EXERCISE 1.5

#### Question 1:

Let

Equating the coefficients of x and constant term, we obtain

A + = 1

2A + B = 0

On solving, we obtain

A = −1 and B = 2

#### Question 2:

Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

#### Question 3:

Let

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

A = 1, B = −5, and C = 4

#### Question 4:

Let

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

#### Question 5:

Let

Substituting x = −1 and −2 in equation (1), we obtain

A = −2 and B = 4

#### Question 6:

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain

Let

Substituting x = 0 and  in equation (1), we obtain

= 2 and B = 3

Substituting in equation (1), we obtain

#### Question 7:

Let

Equating the coefficients of x2x, and constant term, we obtain

A + C = 0

A + B = 1

B + C = 0

On solving these equations, we obtain

From equation (1), we obtain

#### Question 8:

Let

Substituting x = 1, we obtain

Equating the coefficients of x2 and constant term, we obtain

A + C = 0

−2A + 2B + C = 0

On solving, we obtain

#### Question 9:

Let

Substituting x = 1 in equation (1), we obtain

B = 4

Equating the coefficients of x2 and x, we obtain

A + C = 0

B − 2C = 3

On solving, we obtain

#### Question 10:

Let

Equating the coefficients of x2 and x, we obtain

#### Question 11:

Let

Substituting = −1, −2, and 2 respectively in equation (1), we obtain

#### Question 12:

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain

Let

Substituting = 1 and −1 in equation (1), we obtain

#### Question 13:

Equating the coefficient of x2x, and constant term, we obtain

A − B = 0

B − C = 0

A + C = 2

On solving these equations, we obtain

A = 1, B = 1, and C = 1

#### Question 14:

Equating the coefficient of x and constant term, we obtain

A = 3

2A + = −1 ⇒ B = −7

#### Question 15:

Equating the coefficient of x3x2, x, and constant term, we obtain

On solving these equations, we obtain

#### Question 16:

[Hint: multiply numerator and denominator by xn − 1 and put xn = t]

Multiplying numerator and denominator by x− 1, we obtain

Substituting t = 0, −1 in equation (1), we obtain

A = 1 and B = −1

#### Question 17:

[Hint: Put sin x = t]

Substituting t = 2 and then t = 1 in equation (1), we obtain

A = 1 and B = −1

#### Question 18:

Equating the coefficients of x3x2x, and constant term, we obtain

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equations, we obtain

A = 0, B = −2, C = 0, and D = 6

#### Question 19:

Let x2 = t ⇒ 2x dx = dt

Substituting = −3 and = −1 in equation (1), we obtain

#### Question 20:

Multiplying numerator and denominator by x3, we obtain

Let x4 = t ⇒ 4x3dx = dt

Substituting t = 0 and 1 in (1), we obtain

A = −1 and B = 1

#### Question 21:

[Hint: Put ex = t]

Let ex = ⇒ ex dx = dt

Substituting t = 1 and t = 0 in equation (1), we obtain

A = −1 and B = 1

#### Question 22:

A.

B.

C.

D.

Substituting x = 1 and 2 in (1), we obtain

A = −1 and B = 2

Hence, the correct answer is B.

#### Question 23:

A.

B.

C.

D.

Equating the coefficients of x2x, and constant term, we obtain

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

= 1, B = −1, and C = 0

Hence, the correct answer is A.

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